222 lines
8.2 KiB
TeX
222 lines
8.2 KiB
TeX
\section{Universal Constructions for Modules}
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\label{section:universal-module}
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\begin{definition}[Product]
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\label{definition:product-module}
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Let $R$ be a ring and $\seqi{A}$ be $R$-modules, then there exists $(A, \bracsn{\pi_i}_{i \in I})$ such that:
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\begin{enumerate}
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\item For each $i \in I$, $\pi_i \in \hom(A; A_i)$.
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\item[(U)] For any $(B, \seqi{T})$ satisfying (1), there exists a unique $T \in \hom(B; A)$ such that the following diagram commutes:
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\[
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\xymatrix{
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B \ar@{->}[rd]_{T_i} \ar@{->}[r]^{T} & A \ar@{->}[d]^{\pi_i} \\
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& A_i
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}
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\]
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\end{enumerate}
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The module $A = \prod_{i \in I}A_i$ is the \textbf{product} of $\seqi{A}$.
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\end{definition}
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\begin{definition}[Direct Sum]
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\label{definition:direct-sum}
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Let $E$ be a ring and $\seqi{A}$ be $R$-modules, then there exists $(A, \bracsn{\iota_i}_{i \in I})$ such that:
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\begin{enumerate}
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\item For each $i \in I$, $\iota_i \in \hom(A_i; A)$.
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\item[(U)] For each $(B, \seqi{T})$ satisfying (1), there exists a unique $T \in \hom(A; B)$ such that the following diagram commutes
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\[
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\xymatrix{
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A \ar@{->}[r]^{T} & B \\
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A_i \ar@{->}[u]^{\iota_i} \ar@{->}[ru]_{T_i} &
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}
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\]
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\end\{enumerate\}
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The module $A = \bigoplus_{i \in I}A_i$ is the \textbf{direct sum} of $\seqi{A}$.
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\end{definition}
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\begin{proof}
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Let
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\[
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A = \bracs{x \in \prod_{i \in I}A_i \bigg | x_i \ne 0 \quad \text{for finitely many}\ i \in I}
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\]
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For each $i \in I$, let
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\[
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\iota_i: A_i \to A \quad (\iota_ix)_j = \begin{cases}
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x &i = j \\
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0 &i \ne j
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\end{cases}
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\]
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then $\iota_i \in \hom(A_i; A)$.
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(U): Let
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\[
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T: A \to B \quad x \mapsto \sum_{i \in I}T_ix_i
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\]
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then $T \in \hom(A; B)$ and the diagram commutes. Since $\bigcup_{i \in I}\iota_i(A_i)$ spans $A$, $T$ is the unique linear map making the diagram commute.
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\end{proof}
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\begin{proposition}
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\label{proposition:module-direct-limit}
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Let $R$ be a ring and $(\seqi{A}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ be an upward-directed system of $R$-modules, then there exists $(A, \bracsn{T^i_A}_{i \in I})$ such that:
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\begin{enumerate}
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\item For each $i \in I$, $T^i_A \in \hom({A_i; A})$.
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\item For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:
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\[
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\xymatrix{
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A_i \ar@{->}[rd]_{T^i_A} \ar@{->}[r]^{T^i_j} & A_j \ar@{->}[d]^{T^j_A} \\
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& A
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}
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\]
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\item[(U)] For any pair $(B, \bracsn{S^i_B}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom({A, B})$ such that the following diagram commutes
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\[
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\xymatrix{
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A_i \ar@{->}[d]_{T^i_A} \ar@{->}[rd]^{S^i_B} & \\
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A \ar@{->}[r]_{g} & B
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}
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\]
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for all $i \in I$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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Let $M = \bigoplus_{i \in I}A_i$. For any $i, j \in I$ with $i \lesssim j$ and $x \in A_i$, let $x_{i, j} \in M$ such that for any $k \in I$,
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\[
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\pi_k(x_{i, j}) = \begin{cases}
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x &k = i \\
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T^i_j x &k = j \\
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0 &k \ne i, j
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\end{cases}
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\]
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Let $N \subset M$ be the submodule generated by $\bracs{x_{i, j}|i, j \in I, i \lesssim j, x \in A_i}$, $A = M/N$, and $\pi: M \to M/N$ be the canonical map.
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(1): For each $i \in I$, let
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\[
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T^i_M: A_i \to M \quad \pi_k T^i_M x = \begin{cases}
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x &k = i \\
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0 &k \ne i
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\end{cases}
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\]
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and $T^i_A = \pi \circ T^i_M$.
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(2): Let $i, j \in I$ with $i \lesssim j$, then for any $x \in A_i$, $T^i_Mx - T^j_M T^i_j x \in N$. Hence $T^i_Ax = T^j_A T^i_jx$.
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(U): Let
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\[
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S_0: M \to B \quad x \mapsto \sum_{i \in I}S^i_B \pi_i x
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\]
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then $S_0$ is the unique linear map such that $S_0 \circ T^i_M = S^i_B$ for all $i \in I$. For any $i, j \in I$ with $i \lesssim J$, $S^i_B x = S^j_B T^i_j x$, so $\ker S_0 \supset N$. By the first isomorphism theorem, there exists a unique $S \in \hom(A; B)$ such that $S_0 = S \circ \pi$.
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\end{proof}
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\begin{proposition}
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\label{proposition:module-inverse-limit}
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Let $R$ be a ring and $(\seqi{A}, \bracs{T^i_j|i, j \in I, i \lesssim j})$ be a downward-directed system of $R$-modules, then there exists $(A, \bracsn{T^A_i}_{i \in I})$ such that:
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\begin{enumerate}
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\item For each $i \in I$, $T^A_i \in \hom(A; A_i)$.
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\item For any $i, j \in I$ with $i \lesssim j$, the following diagram commutes:
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\[
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\xymatrix{
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A_i \ar@{->}[r]^{T^i_j} & A_j \\
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A \ar@{->}[u]^{T^A_i} \ar@{->}[ru]_{T^A_j} &
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}
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\]
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\item[(U)] For any pair $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2), there exists a unique $S \in \hom(B; A)$ such that the following diagram commutes
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\[
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\xymatrix{
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& A_i \\
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B \ar@{->}[r]_{S} \ar@{->}[ru]^{S^B_i} & A \ar@{->}[u]_{T^A_i}
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}
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\]
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for all $i \in I$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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Let
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\[
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A = \bracs{x \in \prod_{i \in I}A_i \bigg | \pi_j(x) = T^i_j\pi_i(x) \forall i, j \in I, i \lesssim j}
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\]
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For each $i \in I$, let $T^A_i = \pi_i$, then $(A, \bracsn{T^A_i}_{i \in I})$ satisfies (1) and (2) by definition of $A$.
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(U): Let $(B, \bracsn{S^B_i}_{i \in I})$ satisfying (1) and (2). Let
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\[
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S: B \to \prod_{i \in I}A_i \quad \pi_i(Sx) = S^B_i
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\]
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then for any $x \in B$ and $i, j \in I$ with $i \lesssim j$,
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\[
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\pi_j (Sx) = S^B_jx = T^i_j S^B_ix = T^i_j \pi_i(S x)
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\]
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so $S \in \hom(B; A)$, and the diagram commutes. Since any map $f: B \to A$ is uniquely determined by its composition with the projections, $S$ is unique.
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\end{proof}
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\begin{definition}[Tensor Product]
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\label{definition:tensor-product}
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Let $R$ be a commutative ring and $\seqf{E_j}$ be $R$ modules, then there exists a pair $(\bigotimes_{j = 1}^n E_j, \iota)$ such that:
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\begin{enumerate}
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\item $\bigotimes_{j = 1}^n E_j$ is an $R$-module.
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\item $\iota: \prod_{j = 1}^n E_j \to \bigotimes_{j = 1}^n E_j$ is a $n$-linear map.
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\item[(U)] For any pair $(F, \lambda)$ satisfying (1) and (2), there exists a unique $\Lambda \in \hom(\bigotimes_{j = 1}^n E_j; F)$ such that the following diagram commutes:
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\[
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\xymatrix{
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\prod_{j = 1}^n E_j \ar@{->}[rd]_{\lambda} \ar@{->}[r]^{\iota} & \bigotimes_{j = 1}^n E_j \ar@{->}[d]^{\Lambda} \\
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& F
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}
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\]
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\item $\bigotimes_{j = 1}^n E_j$ is the linear span of $\iota(\prod_{j = 1}^n E_j)$.
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\end{enumerate}
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The module $\bigotimes_{j = 1}^n E_j$ is the \textbf{tensor product} of $\seqf{E_j}$, and $\iota: \prod_{j = 1}^n E_j \to \bigotimes_{j = 1}^n E_j$ is the \textbf{canonical embedding}. For any $(x_1, \cdots, d_n) \in \prod_{j = 1}^n E_j$, the image
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\[
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x_1 \otimes \cdots \otimes x_n = \iota(x_1, \cdots, x_n)
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\]
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is its \textbf{tensor product}.
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\end{definition}
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\begin{proof}
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Let $M$ be the free module generated by $\prod_{j = 1}^nE_j$, and $N \subset M$ be the submodule generated by elements of the following form:
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\begin{enumerate}
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\item For any $1 \le j \le n$, $(x_1, \cdots, x_n) \in \prod_{k = 1}^n E_k$, and $x_j \in E_j$,
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\[
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(x_1, \cdots, x_j + x_j', \cdots, x_n) - (x_1, \cdots, x_j, \cdots, x_n) - (x_1, \cdots, x_j', \cdots, x_n)
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\]
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\item For any $(x_1, \cdots, x_n) \in \prod_{k = 1}^n E_k$ and $\alpha \in R$,
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\[
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(x_1, \cdots, \alpha x_j, \cdots, x_n) - \alpha(x_1, \cdots, x_n)
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\]
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\end\{enumerate\}
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(1), (2): Let $\bigotimes_{j = 1}^n E_j = M/N$ and
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\[
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\iota: \prod_{j = 1}^n E_j \to \bigotimes_{j = 1}^n E_j \quad (x_1, \cdots, x_n) \mapsto (x_1, \cdots, x_n) + N
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\]
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then by definition of $N$, $\iota$ is $n$-linear.
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(U): Let $(F, \lambda)$ be a pair satisfying (1) and (2), then $\lambda$ admits a unique extension to a linear map $\ol \lambda: M \to F$. Since $\lambda$ is $n$-linear, $\ker \ol \lambda \supset N$. By the first isomorphism theorem, there exists a unique $\Lambda \in \hom(\bigotimes_{j = 1}^n E_j; F)$ such that the given diagram commutes.
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(4): Since $M$ is the free module generated by $\prod_{j = 1}^n E_j$, $M/N$ is generated by $\iota(\prod_{j = 1}^n E_j)$.
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\end{proof}
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